Is “second-countable implies separable” equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for all $n$, and we may choose some $x_n \in B_n$. Then $D = \{x_n\}$ is easily seen to be dense.

Question This proof needs the Axiom of countable Choice. Can we avoid it? Or is the statement even equivalent to the Axiom of countable Choice? What happens if we also assume that $X$ is a complete metric space?

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Surely this requires only the axiom of countable choice? –  Alon Amit Jan 4 '11 at 18:17
@Alon: Yes :). I've edited this. –  Martin Brandenburg Jan 4 '11 at 18:20
This is form 8L of the axiom of choice at http://consequences.emich.edu/CONSEQ.HTM, and is known to be equivalent to countable choice. The proof is fairly straightforward: if $B_1, B_2, ...$ is a countable collection of nonempty sets, consider the topological space $X$ consisting of the disjoint union of the $B_i$ with the $B_i$ as a base (the partition topology). This space is separable if and only if there is a choice function.